0
$\begingroup$

Let $\mathbb{R}$ be the field of the reals and let $a,b,c \in \mathbb{R}^{\times}$. As you probably know, the Hilbert symbol over any field $K$ is defined as: $$(\frac{a,b}{K}) = 1 \text{ if } \exists x,y \in K \text{ such that } ax^2 + by^2 = 1,$$ and $-1$ otherwise. I've proven the multiplicative result $$(\frac{a,bc}{\mathbb{Q}_p}) = (\frac{a,b}{\mathbb{Q}_p})(\frac{a,c}{\mathbb{Q}_p})$$ for the field $\mathbb{Q}_p$, but can't find the reasoning for $\mathbb{R}$, strangely enough. I have the feeling that it's easily verified, but I'm not seeing it. Any thoughts?

Thanks!

$\endgroup$
2
$\begingroup$

If you work with the real numbers then the existence of $x,y$ satisfying $ax^2+by^2=1$ only depends on wether one of $a,b$ is nonnegative. For example if $a>0$ then $(x,y)=\left(\frac{1}{\sqrt{a}},0\right)$ is a possible solution.

With that in mind it becomes a simple discussion on signs : if $a>0$ then all terms in your equation are $1$, and if not then $\left(\frac{a,b}{\mathbb{R}}\right)$ is the sign of $b$, so the multiplicative property holds.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.